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July 11, 2014

July 11, 2014

Posted by **Julie** on Wednesday, January 21, 2009 at 11:49am.

a. Suppose that on each of the days Monday, Tuesday, and Wednesday five tables are sold. What is the probability that all the tables sold on Monday and Tuesday are plain and that on Wednesday at least one deluxe table is sold?

b. Suppose the number of tables sold each day is a Poisson random variable with parameter x = 5. What is the probability that exactly five tables will be sold on a given day?

- Probability -
**Reiny**, Wednesday, January 21, 2009 at 12:35pmProb(5plain tables) on Monday = .8^5

prob(5 plain) on Tuesday = .8^5

prob(5 plain on Monday AND 5 plain on Tuesday) = (.8^5)(.8^5) = .8^10 = .107

Prob(at least one deluxe) = 1 - Prob(all 5 plain)

= 1 - .8^5 = .672

b. don't know about Poisson random variables.

- Probability -
**economyst**, Wednesday, January 21, 2009 at 1:24pmA poisson distribution follows the formula: (e^(-x) * x^k)/k!

where x is the expected mean and k is the number of "true" events. You are given that the mean number of sales of tables is 5 (x=5) and you are asked for the probability that exactly 5 tables are sold (k=5)

Plug in the formula:

e^(-5) = .066738

5^5 = 3125

5! = 120

poisson=.1755 = 17.55%

See: http://www.answers.com/topic/poisson-distribution

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